\[A_1+A_2 \rightarrow A_3\]
with rate constant \[k_1\]
\[A_3+A_2 \rightarrow A_4\]
with rate constant \[k_2\]
\[A_5+A_2 \rightarrow A_6\]
with rate constant \[k_3\]
\[A_6+A_2 \rightarrow A_7\]
with rate constant \[k_4\]
\[A_4\]
is the desired product. \[A_6, \: A_7\]
are undesirable by products. The problem is to find the minimum number of differential equations to describe the reaction.The rate at which a chemical reaction proceeds is equal to the ratre constant times the concentrations of the reactants. We can use this to find the rate at which the concentration of each chemical in the equations changes. Let
\[x_i\]
be the contraction of chemical \[A_i\]
. A negative sign means the chemical is being consumed by a reaction.\[\frac{dx_1}{dt}-k_1x_1x_2\]
\[\frac{dx_2}{dt}=-k_1x_1x_2-k_2x_2x_3-k_2x_2x_5-k_4x_2x_6\]
\[\frac{dx_3}{dt}=k_1x_1x_2-k_2x_2x_3\]
\[\frac{dx_4}{dt}=k_2x_2x_3\]
\[\frac{dx_5}{dt}=-k_3x_2x_5\]
\[\frac{dx_6}{dt}=k_3x_2x_5-k_4x_2x_6\]
\[\frac{dx_7}{dt}=k_4x_2x_6\]
We can write this in matrix form as
\[frac{d}{dt} \begin{pmatrix}x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \\ x_6 \\ x_7 \end{pmatrix} = \left( \begin{array}{ccc} -k_1 & 0 & 0 & 0 \\ -k_1 & -k_2 & -k_3 & -k_4 \\ k_1 & -k_2 & 0 & 0 \\ 0 & k_2 & 0 & 0 \\ 0 & 0 & -k_3 & 0 \\ 0 & 0 & k_3 & -k_4 \\ 0 & 0 & 0 & k_4 \end{array} \right) \begin{pmatrix}x_1x_2 \\ x_2x_3 \\ x_2x_5 \\ x_2x_6 \end{pmatrix}\]
Reduce the matrix on the right to diagonal form using elementary row operations.
Add -1 times the first row to the second row and add the first row to the third row.
\[frac{d}{dt} \begin{pmatrix}x_1 \\ x_2-x_1 \\ x_3+x_1 \\ x_4 \\ x_5 \\ x_6 \\ x_7 \end{pmatrix} = \left( \begin{array}{ccc} -k_1 & 0 & 0 & 0 \\ 0 & -k_2 & -k_3 & -k_4 \\ 0 & -k_2 & 0 & 0 \\ 0 & k_2 & 0 & 0 \\ 0 & 0 & -k_3 & 0 \\ 0 & 0 & k_3 & -k_4 \\ 0 & 0 & 0 & k_4 \end{array} \right) \begin{pmatrix}x_1x_2 \\ x_2x_3 \\ x_2x_5 \\ x_2x_6 \end{pmatrix}\]
Interchange row 2 and row 3.
\[frac{d}{dt} \begin{pmatrix}x_1 \\ x_3+x_1 \\ x_2-x_1 \\ x_4 \\ x_5 \\ x_6 \\ x_7 \end{pmatrix} = \left( \begin{array}{ccc} -k_1 & 0 & 0 & 0 \\ 0 & -k_2 & 0 & 0 \\ 0 & -k_2 & -k_3 & -k_4 \\ 0 & k_2 & 0 & 0 \\ 0 & 0 & -k_3 & 0 \\ 0 & 0 & k_3 & -k_4 \\ 0 & 0 & 0 & k_4 \end{array} \right) \begin{pmatrix}x_1x_2 \\ x_2x_3 \\ x_2x_5 \\ x_2x_6 \end{pmatrix}\]
Add row 5 to row 6.
\[frac{d}{dt} \begin{pmatrix}x_1 \\ x_3+x_1 \\ x_2-x_1 \\ x_4 \\ x_5 \\ x_5+x_6 \\ x_7 \end{pmatrix} = \left( \begin{array}{ccc} -k_1 & 0 & 0 & 0 \\ 0 & -k_2 & 0 & 0 \\ 0 & -k_2 & -k_3 & -k_4 \\ 0 & k_2 & 0 & 0 \\ 0 & 0 & -k_3 & 0 \\ 0 & 0 & 0 & -k_4 \\ 0 & 0 & 0 & k_4 \end{array} \right) \begin{pmatrix}x_1x_2 \\ x_2x_3 \\ x_2x_5 \\ x_2x_6 \end{pmatrix}\]
Interchange row 3 and row 5.
\[frac{d}{dt} \begin{pmatrix}x_1 \\ x_3+x_1 \\ x_5 \\ x_4 \\ x_2-x_1 \\ x_5+x_6 \\ x_7 \end{pmatrix} = \left( \begin{array}{ccc} -k_1 & 0 & 0 & 0 \\ 0 & -k_2 & 0 & 0 \\ 0 & 0 & -k_3 & 0 \\ 0 & k_2 & 0 & 0 \\ 0 & -k_2 & -k_3 & -k_4 \\ 0 & 0 & 0 & -k_4 \\ 0 & 0 & 0 & k_4 \end{array} \right) \begin{pmatrix}x_1x_2 \\ x_2x_3 \\ x_2x_5 \\ x_2x_6 \end{pmatrix}\]
Interchange row 4 and row 6.
\[frac{d}{dt} \begin{pmatrix}x_1 \\ x_3+x_1 \\ x_5 \\ x_5+x_6 \\ x_2-x_1 \\ x_4 \\ x_7 \end{pmatrix} = \left( \begin{array}{ccc} -k_1 & 0 & 0 & 0 \\ 0 & -k_2 & 0 & 0 \\ 0 & 0 & -k_3 & 0 \\ 0 & 0 & 0 & -k_4 \\ 0 & -k_2 & -k_3 & -k_4 \\ 0 & 0 & k_2 & 0 \\ 0 & 0 & 0 & k_4 \end{array} \right) \begin{pmatrix}x_1x_2 \\ x_2x_3 \\ x_2x_5 \\ x_2x_6 \end{pmatrix}\]
Add -1 times the second row, -1 times the third row and -1 times the fourth row to the fifth row. Add the second row to the sixth row and the fourth row to the seventh row. The result is
\[frac{d}{dt} \begin{pmatrix}x_1 \\ x_3+x_1 \\ x_5 \\ x_5+x_6 \\ -3x_1+x_2-x_3-2x_5-x_6 \\ x_1+x_3+x_4 \\ x_5+x_6+x_7 \end{pmatrix} = \left( \begin{array}{ccc} -k_1 & 0 & 0 & 0 \\ 0 & -k_2 & 0 & 0 \\ 0 & 0 & -k_3 & 0 \\ 0 & 0 & 0 & -k_4 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right) \begin{pmatrix}x_1x_2 \\ x_2x_3 \\ x_2x_5 \\ x_2x_6 \end{pmatrix}\]
There are four independent equations. They are
\[\frac{dx_1}{dt}-k_1x_1x_2\]
\[\frac{d(x_1+x_3)}{dt}=-k_2x_2x_3\]
\[\frac{dx_5}{dt}=-k_3x_2x_5\]
\[\frac{d(x_5+x_6)}{dt}=-k_4x_2x_6\]